there are three consecutive positive integers such that the sum of squares of first integer and product of second and third is 191 .Find those integers. quadric equation
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first write the segment of third is 191, then insert you will draw with a compass
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Answer:
Correct option is
A
4
⇒ Let there consecutive positive integers are x,x+1,x+2.
According to the question,
⇒ x
2
+(x+1)(x+2)=46
⇒ x
2
+x
2
+2x+x+2=46
⇒ 2x
2
+3x+2−46=0
⇒ 2x
2
+3x−44=0
⇒ Here, a=2,b=3 and c=44
⇒ x=
2a
−b±
b
2
−4ac
⇒ x=
2(2)
−3±
(3)
2
−4(2)(−44)
⇒ x=
4
−3±
361
⇒ x=
4
−3±19
⇒ x=
4
−3+19
and x=
4
−3−19
∴ x=
4
16
and x=
4
−22
∴ x=4 and x=
2
−11
⇒ Here, there is only one integer which is 4.
∴ Smallest integer is 4.
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